help-13

[wintry grove]

someone please help me out

<@&286206848099549185>

[wintry grove]

..

[fair geyser]

you're supposed to find area of the rectangles and divide by 2 when it's actually a traingle

[wintry grove]

i'm blank

[fair geyser]

google "area of rectangle" I guess

you probably know that, so maybe "area of right traingle"

[wintry grove]

h*b

ik that

but..

[wintry grove]

like can u explain what u mean by this

[fair geyser]

area of right triangle is (1/2)hb

so it's area of rectangle divided by 2

[wintry grove]

yea that

[fair geyser]

that's what i meant

[wintry grove]

arggh can u just solve it and show I'll understand better

@fair geyser

[fair geyser]

no it's boring

you're supposed to count the squares to find h and b, it's alot of effort, I won;t do it

[crimson sedge]

count the squares

[cedar kiln]

@wintry grove Has your question been resolved?

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[green rose]

Hello good day?

[green rose]

how does 5^x-2 turn into 5^2x?

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[visual plover]

find DNF (pls step by step)

[vale gyro]

well A -> B is equivalent to (not(A) or B)

if you use that, than you have only disjunktions

[visual plover]

thats all?

cos wolfram says its not(r)

[vale gyro]

i think so if the task was the find a disjunktive normel form. That is the defintion of it

[vale gyro]

what is not(r)

?

[vale gyro]

this whole term?

[vale gyro]

a okay, yes that is maybe possible to logic rules. For example r or r is equivalant to r, look at wiki for more rules

another example: not(a or b) = not a and not b

and so on

[visual plover]

okay

thank u

[vale gyro]

np

[visual plover]

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[vestal schooner]

.help

[cedar kiln]

Commands:
clopen: .close, .reopen
factoids: .tag
help: .help

Type .help <command name> for more info on a command.

[vestal schooner]

can someone please explain how these are correct

[vestal schooner]

pls

[radiant topaz]

Wait for helpers to arrive

[vestal schooner]

ok

[regal oak]

mind if I help ?

[vestal schooner]

yes you can

[regal oak]

we know that
76+z=180 ( st line angle = 180)
y+z=180 ( st line angle = 180)
so compare both y = 76

understood ?

[vestal schooner]

yes

is that all?

[regal oak]

ya

similarly z=x

[vestal schooner]

do u think u could explain how z and x are vertical angles pls

[regal oak]

ok vertical angle is the pair of opposite angles formed by intersection of lines

tht is definition

[vestal schooner]

oh so like intercepting lines is what a vertical angle is?

[regal oak]

can u giv a diagram so that it will be ez to explain

[vestal schooner]

wym by that ]

[regal oak]

no drawing for easy explanation

[vestal schooner]

u want me to do what exactly?

[regal oak]

can u provide some diagram so that I can explain you

[regal oak]

x,z and y,76 are vertical angle

[vestal schooner]

do u what m is in this problem?

[regal oak]

m is ment by angle

[vestal schooner]

oh ok

[regal oak]

just a notation

[vestal schooner]

ok

what is the degrees of z and x is that able to found and if how so

[regal oak]

104 (180-76) bcos angle on a line adds up to 180

[vestal schooner]

ok

thank you

[regal oak]

np u can dm if u still get doubt on it

[vestal schooner]

do u have time to help with one or r u busy

[cedar kiln]

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[hallow stream]

.

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[delicate gale]

I do not which rule applies to this because i got inconclusive from the ratio test

[delicate gale]

Also cant figure the limit with divergence test

[faint dirge]

as a rule of thumb, exponentials grow faster than polynomials, so it converges

but the ratio test should work?

$\lim_{n->∞}\frac{\frac{(n+1)^2}{5^{n+1}}}{\frac{n^2}{5^n}}$

[wraith dagger]

Scythe

[faint dirge]

$\lim_{n->∞}\frac{\frac{(n+1)^2}{5}}{n^2}$

[wraith dagger]

Scythe

[faint dirge]

$\frac{1}{5}\lim_{n->∞}\frac{(n+1)^2}{n^2}}=\frac{1}{5}\lim_{n->∞}\frac{2(n+1)}{2n}}=\frac{1}{5}\lim_{n->∞}\frac{2}{2}}=\frac{1}{5}$

[wraith dagger]

Scythe
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[faint dirge]

since the ratio is less than 1 it converges, no?

[delicate gale]

i see, i acccidently skipped a step leading me to get 1/1, thank you very much

[faint dirge]

glad I could help :D

[delicate gale]

thanks again

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[terse osprey]

hey, i need someone to please verify my process here.
the way i see it i'm supposed to solve it this way:
define v as the projection of u on the system.
so
v=Σ<u,e_n>e_n=Σ(n)^(-1/3)e_n
now since ║v║≤║u║
║Σ(n)^(-1/3)e_n║≤║u║
and so here's where i'm uncertain.

i think i'm supposed to arrive at ║v║=Σ n^(-1/3)
this series diverges and so v (and u) cannot exist in V

my issue is with getting to ║v║=Σ n^(-1/3)
im not sure how i work this out when i don't have an explicit inner product to evaluate it.
i think i'm really close but im missing a tiny step and im derping.

[terse osprey]

i mean is it:
║v║= sqrt(<v,v>)= sqrt((Σ(n)^(-1/3))^2)

by IP such that w=(w1,w2...wn) <w,w>=w1·w1+w2·w2+wn·wn

why is that necessarily true if so?
how do i know it's this IP and not some other IP?
i feel like i'm stumbling on something rly basic

<@&286206848099549185>

[cedar kiln]

@terse osprey Has your question been resolved?

[terse osprey]

<@&286206848099549185>

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[fading narwhal]

try drawing a graph of the line first

[crimson sedge]

Cant it be done in some other way...?

[fading narwhal]

I think drawing a quick sketch would be easier since you can visualize it

[brave aspen]

You dont need a graph at all

Use formula for midpoint

X intercept has coordinates: (a,0)

Y intercept has : (0,b)

Mid-point coordinates are the averages of the coordinates of the 2 points

So

MP (a/2 , b/2) = (2, -5)

So a/2 = 2.

b/2 = -5

A = 4, b = -10

P is the y intercept so its (0, -10)

[crimson sedge]

Can you do this on notebook or @wraith dagger

[brave aspen]

Like on paper?

[crimson sedge]

Yeah

[brave aspen]

@crimson sedge

[crimson sedge]

Ty

I see now

Ooh got it now...🤔🤔@brave aspen

Tysm

[brave aspen]

Yw

[crimson sedge]

@fading narwhal yoi 2

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[ashen olive]

Hi there

[ashen olive]

Need some help in Linear Algebra

I am in 3.b now,

I was wondering about that

how to start

[dense wing]

Hint: consider elementary matrices

And part a

[ashen olive]

by the way

about a

proving it with their hint

was just going back to the definition of REF matrix

isnt it?

I mean it felt like the answer was too quick but idk

maybe I was wrong?

[dense wing]

Pretty much from what I can see

[brave aspen]

If theyre asking you to prove it you have to do it for a general square matrix

With nxn dimensions

Not a specific size of matrix

[dense wing]

What

Where are you getting they used a set size?

[brave aspen]

Im not but to say the answer was quick makes me think they did not do it for a general matrix

Just offering a caution

[dense wing]

I mean... the hint QED

It's just a fast proof lol

[ashen olive]

I proved for a general square matrix

[brave aspen]

Ok great

I just suck xD

[dense wing]

But yeah, try B with the hint I gave

[ashen olive]

well I know that an elementary matrix is invertible, and I know that a square matrix is invertible iff A is row-equivalent to I

I know that PA has to be REF,

which means multiplying A by P from the right, will make A a REF matrix

Oh

I know that A is invertible so it is a product of elementary matrices

but I still don't know how o

to start with this one..

@dense wing

<@&286206848099549185>

anyone??

@dense wing

[cedar kiln]

@ashen olive Has your question been resolved?

[dense wing]

What does multiplying by an elementary matrix do?

[cedar kiln]

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[crimson sedge]

So my dy/dt is 1/(e^(4+ln2) + 1)

Then I found dt/dx

4 e^(4+ln2)

[glad kestrel]

there is no variable in your equation

derivative is 0

[crimson sedge]

What do you mean

[dense wing]

You don't have x

[glad kestrel]

ln(e^(4+ln(2))+1= a number

derivative of a number = 0

[crimson sedge]

Oooooh

Thanks

That was way easier than I thought

Quick question for dy/dx(ln(ln(ln(x)))) will I have to use the chain rule more than once?

[royal musk]

Yes

[crimson sedge]

Thanks

[cedar kiln]

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[crimson sedge]

Another question

[cedar kiln]

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[wise moat]

there was a mistake here right?

the delta x1 should be squared

it's in the textbook

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[crimson sedge]

Can someone help me with this problem?

[crimson sedge]

im suppose to use a calculator for this problem but idk how to start it

I got to cosx = (x+1)(x-1)

but idk if thats even right

[shell cloud]

Set it as a y = equation

And then graph

[floral thistle]

good thing i didn’t say newtons method

[shell cloud]

What's Newtons method?

[floral thistle]

root finder

[graceful karma]

[floral thistle]

involves derivatives

[shell cloud]

Ohh I see

Linearization right?

[floral thistle]

not sure what you mean by that honesty

[shell cloud]

I think it's the same thing I just learned it as "linearization" for the approximation of points

[floral thistle]

you mean linear approximation?

[shell cloud]

Yeah

So I guess I'm thinking of the wrong thing then

[cedar kiln]

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[crimson sedge]

.close

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[kind venture]

what would be the best way to come up with all pairs (dice 3 times) that sum of to a specific number for example : (1,2,5) = 8...find alll tulpes that give us the total sum of 8 ?

[kind venture]

like is there a systematic way to find them

maybe by making a table or smth

for example here... dice two times and get sum = 8...

this is how I would find all pairs

(2,6) = 8... and so on

how would I do that for 3 numbers ?

[cedar kiln]

@kind venture Has your question been resolved?

[tranquil remnant]

x + y + z = w, 0<x,y,z<7, 2<w<19

you would need a 3 dimensional paper to do the table for 3 numbers

@kind venture ^

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[ashen olive]

Hi

[ashen olive]

I have another question in linear algebra

anyone can help ?

[tranquil remnant]

If the sum of all rows is zero, then the rows are linearly independent. So the determinant is zero.

[cedar kiln]

@ashen olive Has your question been resolved?

[ashen olive]

But isn’t it the sum of all columns?

@tranquil remnant

[tranquil remnant]

J is the one we fix, and I is the one we continuously change

Is there any other info on the question?

Like are all entries positive?

[ashen olive]

Nope

no other infor

information

but if j is the fixed one

and I is the one we change

[tranquil remnant]

Yeah you were right

it is columns that add to zero

Ah ok

yeah, since all the elements in each colum add to zero, if we add al rows we get the zero vector

[ashen olive]

hmmmm

how do I write it?

like

[tranquil remnant]

and thus the rows are not linearly independent, so the determinant is zero

[ashen olive]

We did learn that if we are a zero row in the matrix the determinant of it is 0

but we haven't learned about the connection between the sum / 2 or more similar rows to the determinant being 0

youre saying that if we have any identical rows, we get determinant of zeo?>

zero?

[tranquil remnant]

take column one; i.e. (X11,X12,...,X1N)

up until column N; i.e. (XN1,XN2,...,XNN)

if you add them, youll have: (X11+...+XN1,...,X1N+...+XNN)

which is equal to (0,...,0)

so the row vectors are linearly dependent, hence the determinant is zero

[ashen olive]

I do get the way of doing it

I don't think I get the defenition of linearly dependent, and the reason to do it

if I want to calculate the determinant

ill just do It with the formula

who said I'll get 0?

[tranquil remnant]

yeah, but you don't have to. If you know that there is a non-trivial linear combination of rows, which is equal to the zero row, then you have that the determinant is zero

[ashen olive]

yeah.. sorry for my question

it is just that we haven't learned that yet

you see? I can't just write that

I am trying to write that correctly

idk if thats right

let me send you what i have done

i think it is not ok to write like I did

@quaint silo

[tropic oxide]

dependent*

[ashen olive]

Ann

[tranquil remnant]

indeed

thanks

[ashen olive]

what is did is not good?

@quaint silo

@tropic oxide

I mean

Is there a way to say that?

I dont know if I can use Sigma for that...

I dont get it

I am sorry

I really don't understand what you say

[dense wing]

Assume A != 0
For any j, that column's entries will sum to 0.
Taking transpose of A maintains determinant, so consider det(A^T).
In A^T, the row entries now sum to 0.
Hence if you vector add all the columns in A^T, you get the 0 vector

[ashen olive]

det (A^T) = det(A)

?

[dense wing]

Yes.

[ashen olive]

nice

we haven't learned that yet tho

[dense wing]

Ok, then don't use it

The sum of rows of A give the 0 vector

[ashen olive]

yup

thats what I did

[dense wing]

Hence the vectors span to 0 non-trivially

[ashen olive]

I just wasnt sure how to write that

[dense wing]

"The sum of the rows of A gives the 0 vector"

You're allowed to use words...

[ashen olive]

But it sums to a row of [0,0,.......,0] (n times)

[dense wing]

yeah... and?

[ashen olive]

oh

you mean a vector 1xn

yeah

[dense wing]

Vectors add to vectors, yes

[ashen olive]

I am just not used to see it like that

may I ask one more question?

Do you have a hint about this one?

one where to start?

on*

[dense wing]

it's 2x2 matrices, just write a general one

and compute RHS

[ashen olive]

What is RHS?

[dense wing]

right hand side.

[ashen olive]

is that a technique?

for doing something?

[dense wing]

...

do you know your left and rights?

Just compute the RHS of the formula you're proving

[ashen olive]

oH

Oh

sorry

we don't say that in hebrew

I just did not understand for a moment

[cedar kiln]

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[clear grail]

how do i solve this?

[livid hound]

start with suvat equation(s)

[cedar kiln]

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[cedar kiln]

@orchid flower Has your question been resolved?

[cedar kiln]

@orchid flower Has your question been resolved?

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@orchid flower Has your question been resolved?

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[formal drum]

hi

[crimson sedge]

four vectors

[formal drum]

is there a linear transformation L so that L(u1) = u2 and L( u3) = u4, and that L transforms the plane −x −y + z = 0 on x + z = 0?

so the answer to the first part of the question is yes there are infinite but i need some assurance that the answer to the whole question is no

L(x) = Ax i have an expression for A from proving the first part of the question

X11, X21 and X31 can be chosen freely

[cedar kiln]

@formal drum Has your question been resolved?

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@formal drum Has your question been resolved?

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@formal drum Has your question been resolved?

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[rugged matrix]

what does this mean? I've never seen this before

[obsidian coral]

Which part exactly?

[rugged matrix]

like why is dx there

[obsidian coral]

That's not dx, that's delta

The Greek letter, $\delta$

[rugged matrix]

well yeah

[wraith dagger]

dldh06

[rugged matrix]

i get that

but surely it just evaluates to 0

is it a misprint? like was it meant to be an integral instead of a sum

oh nvm, its just an integral but written differently

.close

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[cedar kiln]

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[queen adder]

@obsidian coral did i do it right? And what do i do after this

[obsidian coral]

Looks good

[queen adder]

its for hyperbola

i think i remember hearing that the number in the right should be 1?

@obsidian coral

so should i divide both groups by 30 to make the 30 in the right 1?

[obsidian coral]

Yes, sounds right

So you divide by 30

To all the terms

[queen adder]

yeah

So this?

[obsidian coral]

You can simplify the left side, but yes

[queen adder]

i havent done math in a while so uh

3 becomes 1 and 30 becomes 10?

[obsidian coral]

What's $\frac{-2}{30}$ and $\frac{3}{30}$?

[wraith dagger]

dldh06

[queen adder]

i simplify it right?

[obsidian coral]

Yes, simplify

[queen adder]

1/-15 and 1/10?

[obsidian coral]

Yes

[queen adder]

aight thanks alot

.close

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[faint sun]

Need help with this

[faint sun]

Not sure where to begin

[obsidian coral]

$Z score = \frac{\bar{x} - \mu}{\sigma}$

[wraith dagger]

dldh06

[faint sun]

Mhmm

[obsidian coral]

Does that equation ring any bells?

[faint sun]

Nope

Ik abt the z score table tho

[obsidian coral]

So there is an equation, the above one, to find the Z score

You are trying to find $\bar{x}$

[wraith dagger]

dldh06

[faint sun]

Yea

[obsidian coral]

So the first part, you need less than 82.4, so use the z table, and find the corresponding z score that is the closest to 82.4 that doesn't go over it

[faint sun]

Can’t find it

Don’t see a 82.4

[obsidian coral]

Don't forget that's 82.4%, you want a decimal

And you want z score that is closest to that value

Here is hint, it's within this block

[faint sun]

So far i did

82.4-90/2.176

And I got -3.492

Idk what’s next

[obsidian coral]

No, 82.4 is not $\bar{x}$

[wraith dagger]

dldh06

[obsidian coral]

82.4 is the percent for the z score

Find the z value that corresponds to 82.4

[faint sun]

What do I do

I’m lost now

[obsidian coral]

Do you know how to use z table?

[faint sun]

Sort of

[obsidian coral]

Like given z value, to find the percentage?

[faint sun]

Nah

[obsidian coral]

Because the z score, corresponds to a percent area under the curve

[faint sun]

Yea

[obsidian coral]

So z score of 0 means 50% of the area under the curve

So you are working backwards, you are given the percent, find the z score

[faint sun]

So what is my z score

[obsidian coral]

You use the z table

[faint sun]

And look for 2.176?

[obsidian coral]

No

You look for 82.4% or a value close to 82.4% but not over

That's the z score

[faint sun]

How do I find that score idk what to calculate to look for it on the table

[obsidian coral]

Let me give you an example. Say you were given 60%, that's 0.6000, use the screenshot I gave, closest value to 0.6000, without going over 0.6000, is 0.5987, on the table, 0.5987 corresponds to a z value of 0.25

[faint sun]

I got -0.0572

82.4-90/17sqrt61

[obsidian coral]

No

That's 82.4%

Use my example as help

[faint sun]

Says less than 82.4

I don’t see percent

[obsidian coral]

It means 82.4%, the question gave you the percent area under the curve

[faint sun]

Sorry bro but you just have me lost

[obsidian coral]

Even though the problem you are asking just says 82.4, it means 82.4 percent

Do you know what the percent means for a normal distribution curve?

The area under the curve?

[faint sun]

Nope

I learned all of this thru this server so far

Could you solve this question so I can apply the same method to the following 2 remaining questions

[obsidian coral]

No

You are lacking that knowledge, you should look at resources on z score and z table and normal distribution curve

[faint sun]

Okay thanks

Have a great night/morning

[cedar kiln]

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[serene tendon]

need help on this question

[cedar kiln]

@serene tendon Has your question been resolved?

[serene tendon]

<@&286206848099549185>

[radiant topaz]

That's a lot of spaces

[serene tendon]

yeah but once u get how to do it its not that bad

im just having trouble understanding what it supposed to go for the equation

F. dA = integral from a to c and integral from c to d = ()

[radiant topaz]

6y+10z

[serene tendon]

i thought the point here is to calculate flux for each variable seperately

oh ok

that would explain the dz dy part

[radiant topaz]

At x=3 , F(3,y,z) = what

[serene tendon]

but then where does x = 3 mean

oh

[radiant topaz]

And it's div

Not grad

[serene tendon]

u meant gradient?

[radiant topaz]

Nope

[serene tendon]

also it's not 6y+10z

[radiant topaz]

Ok wait

[serene tendon]

yeah im really confused because first i tried 10x+6y+10z, that was incorrect, then i tried 5x^2, that also didn't work

[radiant topaz]

The plane x=3 right so normal to that plane is (1,0,0)

Try 10x

[serene tendon]

i tried 10x before, also doesn't work

[radiant topaz]

-10x because the direction of normal inside bounds will be (-1,0,0)

[serene tendon]

still wrong

[radiant topaz]

Now even -5x²

[serene tendon]

still wrong loool

[radiant topaz]

Ok wait I need to solve it instead of assuming

F.n = 5x² tho

[serene tendon]

yeah thats what im confused with as well

[radiant topaz]

Plus (1,0,0) is a unit vector

So what's the problem

[serene tendon]

does this maybe work?

when x = 3

normal changes or something

[radiant topaz]

The figure is like a cube so you can imagine the normals

Or cuboid

Make one attempt on -3y²-5z²

[serene tendon]

ok so something weird happened

i assumed 5x^2 was correct and subbed in x = 3

so 5*(3)^2 * 2 * 1 (2,1 are bounds for the integration) and got = 90

and 90 is correct

but that doesnt make sense, why wont 5x^2 work then

[radiant topaz]

Wait do 45

[serene tendon]

-10 worked as well for the next

45 in the first box?

oh it worked

LOL so it wants a value instead of the expression?

[radiant topaz]

F(3,y,z) . (1,0,0) dydz it is

[serene tendon]

ok so this is what i have

[radiant topaz]

How did you do it so fast

[serene tendon]

i kind of just understood the pattern

but for those 2 red boxes

idk what its supposed to be because

dz dx idk what is

i thought it would just be dz * dx

or maybe its 3-1 * 2-0

lemme try that

I GOT IT

ok i dont really understand how that multiplier works but i caught onto the pattern

so

dx,dy,dz are their respective bounds

so you multiply by the bounds and naturally, if the lower bound isn't 0

you have to do upper - lower

so 3-1 = 2

i was just doing upper x upper because the first one had both lower bounds of 0

thats why it didnt work

for the last ones

but still thats so wack, why doesn't 5x^2 just work lol

are u there boss

@radiant topaz

[radiant topaz]

Yes I'm here reading

Got pinged in other channel too

[serene tendon]

oh i see

well, i only have 1 more question left for my assignment that im struggling on. just the 2nd part i need help

how do i get dr here

[radiant topaz]

What happened

[serene tendon]

so im just struggling on how to get dr and F lol

how do i start to do this

[radiant topaz]

Maybe I can help you later I gtg now

[serene tendon]

oh ok, i gotta submit it by 11:59 so hopefully someone else can help me then haha. Take care

thanks for your help on the first one

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

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@serene tendon Has your question been resolved?

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[oak summit]

Find the formula for each

[oak summit]

I just cant picture it together :(

Formula to find area

[cedar kiln]

@oak summit Has your question been resolved?

[short blade]

do you know the formulas? like do you know the formula for the area of a triangle?

[cedar kiln]

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[rotund night]

would someone be able to teach me how to do this word problem? It is in the conics unit

[rotund night]

@steady imp I saw ur message earlier, i was just trying to figure it out

and i got it

thank you!

.close

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[ashen gale]

On the first day I traveled 112km, on the second day I traveled 3/7 of the first day's distance. Now 7/11 of the total distance is left to travel. What is the total distance?

[final pendant]

let the total distance be x

use this

[ashen gale]

112+112*3/7=4/11x

[final pendant]

yes

(4/11)*x

[ashen gale]

Aye

[final pendant]

important

[ashen gale]

this is like 4th grade problem btw lol
Got da answer though thx

.close

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[lethal goblet]

Suppose y varies inversely as x^4.

Find the percentage change in y if x is decreased by 2%.

[lethal goblet]

Please help with question, i dont understand how to solve it

[final pendant]

what do we mean by "inverse proportion"

when we say y varies inversely as x^n

[lethal goblet]

hmmmm

sorry

my wifi had a problem

i dont know y increases and x decreases?

[final pendant]

yes

basically

y=c*(1/x^n)

where c is some constant

[lethal goblet]

okay

[lethal goblet]

x is 102% now so x would be (1.02x)^4?

.CLOSE

.close

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[chrome quail]

can somebody help me with two algebra 2 long division questions

[chrome quail]

the question is (16x^3 + 4x^2 + 2x - 7) div by (4x-1)

i was reviewing my notes and i tried to plug the steps in but what i got didnt make sense

and ik you can use an app to get the answer but i need to show my work

and i cant find an app that has all the work

do you know any website i could use

[glass jungle]

can you chuck your working here?

[chrome quail]

that would give me the answer to a long division problem

[glass jungle]

you shouldnt rely on those sites :c

[chrome quail]

yes i agree

but i cant find the error in my work

[glass jungle]

if you send your working here i can take a look

[chrome quail]

ok thank you one second

that pic isnt great should i send anohter one

another

any ideas?

[glass jungle]

im kinda rusty on polynomial division but what was your first step?

not sure if we're using the same methods

[chrome quail]

i divided the leading term

16x^3 and 4x

[glass jungle]

mhm

you should chuck your answer above 16^3

and then after that what did you do

[chrome quail]

i got 4x^2 so i multiplied that by x^2 -1

[glass jungle]

why by x^2 -1

where'd you get that from

[chrome quail]

im checking

[glass jungle]

you should checkout how they did it in the Example section of this and apply it to your thing, see if that helps but if not, feel free to ask more
https://www.mathsisfun.com/algebra/polynomials-division-long.html

[chrome quail]

oh i think i see my mistake

it should be 4x^2 multiplied by (4x-1)

[glass jungle]

mhm

sorry but i gtg, hopefully that helps!

[radiant topaz]

Yes the first term of quotient

[chrome quail]

can i send one more problem that i already completed for someone to check?

would truly help me

[cedar kiln]

@chrome quail Has your question been resolved?

[radiant topaz]

Yes

[cedar kiln]

@chrome quail Has your question been resolved?

[cedar kiln]

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[mellow merlin]

Let $p>0$ be a fixed parameter. To each starting value $a_{0} \in \mathbb{R}$, we define a sequence $\left(a_{n}\right){n \in \mathbb{N}}$ recursively with $a{n+1}:=a_{n}\left(2-p a_{n}\right)$. For which starting values $a_{0}$ does the sequence converge (determine the limit) and diverge to $\pm \infty$? \
What I have so far is:
If the sequence does converge, then the only possible limits are $a=0$ or $a= \frac{1}{p}$.\
For $a_0 < 0$ and for $a_0 > \frac{2}{p}$, the sequence diverges to $-\infty$. \
For $a_0 \in \left(0, \frac{2}{p}\right)$, the sequence converges to one of the limits above.\
For $a_0 = 0$ and $a_0 = \frac{2}{p}$, it's obvious.

[wraith dagger]

lewis

[cedar kiln]

@mellow merlin Has your question been resolved?

[mellow merlin]

<@&286206848099549185>

[cedar kiln]

@mellow merlin Has your question been resolved?

[cedar kiln]

@mellow merlin Has your question been resolved?

[dry bane]

In the rectangle below the width is x cm and the length is 5 cm greater than double the width.

how to find the perimetre

[cedar kiln]

@mellow merlin Has your question been resolved?

[viral cloak]

not your channel

[mellow merlin]

not like anybody's gonna help 😔

[cedar kiln]

@mellow merlin Has your question been resolved?

[cedar kiln]

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[bold hinge]

i asked this before in #calculus channel but didn't get an answer

[bold hinge]

take the rectangle with corners in R2 at (0,0), (4,0), (0,5), (4,5)

how do you evaluate the area A as a double integral of x and y?

[leaden otter]

$\int_0^5\int_0^4,dx,dy$

[wraith dagger]

Euclid31415

[bold hinge]

it's just dx and dy?

i put xy dx dy

[leaden otter]

Yes it’s just dx dy

[bold hinge]

when would you see xy dx dy then? does that only apply in R3?

[leaden otter]

That would apply if you are integrating the function f(x,y)=xy over your region

In our case, we are integrating f(x,y)=1 over the rectangle region

[bold hinge]

ok so let me get this straight:

the double integral in R2 is the projection of f(x,y) in R3 to the xy plane

like the "shadow" of the surface formed by the f(x,y) onto the xy plane

?

[leaden otter]

Well, it is a bit of like “weighted” shadow

Where the weight is the function evaluated at that point

[bold hinge]

huh?

[bold hinge]

how do you weight a shadow

[leaden otter]

You can think of it like plotting a curve in 3D and calculating its volume restricted by your region in x-y plane

[bold hinge]

ok so it's just the projection onto the xy plane

?

[leaden otter]

The region you are integrating over is present in the x-y plane

[bold hinge]

if you have a diagram of some sort that would be helpful

or maybe just a link. trying to teach myself calc 3 haha

[leaden otter]

This is f(x,y)=z plotted in 3D

Volume of the region between the curve and the x-y plane is given by\$\int_c^d\int_a^b f(x,y),dx,dy$

[wraith dagger]

Euclid31415

[bold hinge]

OHHHHH

it's VOLUME

not area

[leaden otter]

It is area if you choose f(x,y)=1

[bold hinge]

wow that makes a lot more sense

it's just like a single integral.

in R2 you take the area under the curve. this is just the volume under the curve

[leaden otter]

Basically yes

Yep

[bold hinge]

sweet thanks

[leaden otter]

We loose a bit of visualisation when we have multiple integrals

It is helpful to think of “weights” in that case

[bold hinge]

like a triple integral would require 4 dimensions? w = f(x,y,z)

[leaden otter]

You can’t really plot it

You could encode it like a density value at a particular point in 3D

[bold hinge]

oh ok that sort of makes sense

i was confused when you brought up weights at first

[leaden otter]

If we are integrating over where the function goes below the x-y plane, our volume will be counted as negative

[bold hinge]

yeah checks out

it's pretty much the same thing haha

one more question if you don't mind

how would you set up the integral for the area of this surface? i guess that's analogous to the 2D arc length

@leaden otter ?

[leaden otter]

Well, it would be written in terms of partial derivatives

$\iint_{R}\sqrt{f_x^2+f_y^2+1},dA$

[wraith dagger]

Euclid31415

[leaden otter]

R is the region we are integrating over, dA is the infinitesimal piece of area like dx dy, f_x is the partial derivative wrt x, f_y is the partial derivative wrt y

[bold hinge]

f_z = 1 so yeah that checks out

does this work out in terms of units?

you are integrating sqrt(meters^2) = meters

wait no it's not meters^2?

oh nvm. it's just:
integral of number wrt Area = number * Area = Area?

[leaden otter]

Yep

Think of the 2D analogue

[leaden otter]

Well, partial derivative wrt z doesn’t make a lot of sense in case of f(x,y)

[bold hinge]

dz/dz = 1

$\int_a^b \sqrt{1 + f'(x)^2} dx$

[wraith dagger]

EndTimes

[bold hinge]

right? so there's my final question for you. in this integral we have 'dx'. and that is a variable that is present in the integrand

where is A?

none of the terms are in terms of A

so how do we get dA?

[leaden otter]

dA is the infinitesimal piece of area

[bold hinge]

dA = dy dx?

[leaden otter]

Well, I think yes

[bold hinge]

yeah that makes a lot more sense

so really the formula should be:

$\iint_{R}\sqrt{(\frac{dz}{dx})^2+(\frac{dz}{dy})^2+1} dx dy$

[wraith dagger]

EndTimes

Euclid31415

[bold hinge]

right my bad

it's a partial derivative

do you ever use 'd' at all then?

[leaden otter]

dx dy isn’t the only option. There is also integral in terms of polar coordinates for instance

[bold hinge]

so d is only for single variable and integrals because it's the actual infinitesimal change in variable

humor me once more

wait how do i write swirly d

[leaden otter]

Partial derivatives are there to signify that all other variables are considered as constants wrt the variable you are differentiating

$\partial$

[wraith dagger]

Euclid31415

[bold hinge]

i have too many questions

i don't want to keep you going forever though

[leaden otter]

I also have to go

[bold hinge]

alright thanks

[leaden otter]

Read some sources

[bold hinge]

you don't have to respond to this i'm just writing this down for the sake of it

$\int f(x) \partial x$

[wraith dagger]

EndTimes

[leaden otter]

Cursed

[bold hinge]

yeah i figured

.close

[cedar kiln]

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[blissful mica]

hey, I think my answer is correct but I still don't know why in the formula they put 9 if it is supposed that in that space the first number should be there, in this case 6 so I'm confused. This

[livid hound]

all answers are (simplified)
also show your work

[blissful mica]

(srry if I took too much, I didn't know how to upload an image from my phone) I used both formulas but still don't get the anwer

sorry again for my broken english btw

[livid hound]

,rotate

[livid hound]

you didn't respect the order of operations

also missing some ) on the end

note that
(n-1)d represents the the product of (n-1) and d

and when you substitute d=-3,
you should have something that represents the product of (n-1) and -3

which you do not

[cedar kiln]

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[crimson sedge]

I need help answering this question i tried many times and only got 1 right.

[crimson sedge]

parabolas

@wraith blaze

you have to use these to match it with the vertex

like drop and place

you have to figure out which of these answers

go into the box

im confused.

its parabolas

and putting the coordinates to match the vertex

but im having trouble matching the coordinates with the right vertex

oh

by each box if you get it wrong or right

let me show u

[livid hound]

i think they're trying to ask if this is a test

[crimson sedge]

no

its not

its a workbook

assignment

see

[livid hound]

ok. forget about the answer options for now

lets start with the first one

can you identify/mark the location of the vertex of this parabola

[crimson sedge]

ive never been good at coordinates

so not really

[livid hound]

not asking about the coordinates atm

just mark the location with a marker in an editiing/paint tool

[crimson sedge]

the red dots?

[livid hound]

well there are 3 red dots,
which one do you think is the vertex

[crimson sedge]

the third one?

[livid hound]

if by third one you mean:
then no, that is not the vertex

[crimson sedge]

i have but

im just like on crunch time right now

to get it done

the material is hard to read aswell

its handsoff learning

so im just me without teachers

they just give us the assignment

[livid hound]

if you've read up on the vertex of a parabola, identifying the vertex on the parabola should be relatively simple and intuitive

[crimson sedge]

im listening

[livid hound]

are you able to identify/draw the axis of symmetry for the above parabola?

[crimson sedge]

sorta I struggle

[livid hound]

try it

and show me what you drew

[crimson sedge]

I have to go

but thank you for your guys time

im sorry i hope i'll be able to answer this by myself

thank you.

.close

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[quartz sedge]

find all solutions of sec^2x-sec x = 2 in the interval [0,2pi)

[flint hinge]

you can express this as a quadratic in sec(x)

saying sec(x) = t

we have t^ - t = 2 or t^2 - t -2 = 0

this has 2 solutions viz t = 2 and t = -1

then you can just solve for the corresponding angles

[quartz sedge]

ok let me try that

so it would be pi/3, pi, and 5pi/3?

[flint hinge]

yes!

[quartz sedge]

Ok thanks for your help

[flint hinge]

np

[quartz sedge]

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[lilac bison]

say i have x^2 = y. i know that solving x will be x = +-sqrt(y). In this pic i tried solving for x by taking both (positive and negative) square roots of both sides. is this correct? i'm sure it's not since i basically got x = sqrt(y). i'm just a little confused on how to solve for x. what are we doing to both sides of x^2 = y to solve for x? in x = +-sqrt(y), it seems like first we're taking the principal square root of both sides, but then we're only multiplying the rhs by -1 and leaving the x alone.

[indigo zenith]

Could some one help me at II

[cedar kiln]

@lilac bison Has your question been resolved?

[lilac bison]

<@&286206848099549185>

[brave aspen]

@lilac bison

Consider square root of 4.

Of course 2 is an answer 2^2 =4

But so is (-2)

(-2)^2 = (-2)*(-2) = 4

When you square something it becomes positive so there are two answers for x^2 = y

The positive version of sqrt( y ) and the negative version

It's true for any positive real number

And for all numbers

But idk if you're doing complex numbers

[lilac bison]

i'm only working with real numbers rn

but i can't do the same thing to both sides?

like as what i said

[elfin hemlock]

you can take the sqrt on both sides, but the +- shouldn't be there on both

[livid hound]

$x^2 = y \
\sqrt{x^2} = \sqrt{y} \
|x| = \sqrt{y} \
x = \pm\sqrt{y}$

[wraith dagger]

ℝamonov

[lilac bison]

i see

thanks a lot guys

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[red schooner]

I'm not sure if I understand this correctly. To me this seems like just an overly complicated way of saying something that's intuitively quite obvious
So I suspect that there might be more to it than how I interpret it

[red schooner]

Doesn't it just say "since a normalized distribution sums to one, it has a gradient of 0"?

Why do you need half a page of convoluted equations to derive this?

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[slate fox]

Is there something like Markov chains with weights? Suppose I have very long chains of AA...ABB...BAA....A, then the diagonal of probability matrix will be close to 1, which is not practical

[slate fox]

Is there a way how to solve this issue?

Suppose from intervals (A, 50), (B, 34), (A, 12), ...

[cedar kiln]

@slate fox Has your question been resolved?

[cedar kiln]

@slate fox Has your question been resolved?

[tropic oxide]

your question is unclear

[cedar kiln]

@slate fox Has your question been resolved?

[slate fox]

In what way unclear?

It is intuitive that if the diagonal consist of numbers close to zero, than it is difficult to extract a meaningful information from off diagonal elements

[tropic oxide]

how so?

what information are you looking to extract that's hampered by the transition matrix being diagonal-heavy?

[cedar kiln]

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[dense wing]

what

if the opposite side doubles... then the opposite side gets doubled

???

just arcsin(2*opp).

[bold hinge]

try drawing a picture

[dense wing]

where I'm assuming you mean a hyp of 1.

Yeah... it does in fact work.

[cedar kiln]

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[bold vine]

You can google that

[vale gyro]

.close

[floral thistle]

46656

[leaden otter]

,w 6×6

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by Euclid31415#9927

[floral thistle]

stop

i wish i had helper perms but not the role

for this exact reason

@calm sierra

close the channel please

[calm sierra]

@empty falcon knock it off

.close

[cedar kiln]

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✅

[floral thistle]

@calm sierra sorry for another ping

[calm sierra]

keep trolling & you're out

[floral thistle]

should i just ping mods next time? i only pinged you because i knew you were active right now

[eager elk]

lul

[wooden maple]

[calm sierra]

b&

generally a modping works

.close

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[tropic oxide]

$\sqrt{x^2} - 1$?

[wraith dagger]

Ann

[tropic oxide]

then you should have said sqrt(x^2 - 1)...

okay, so what is troubling you here?

...wait, ok, so you had a test and this problem showed up, yes?

do you have your answer written down somewhere?

you wrote $\frac{1}{2} \sqrt{x^2 - 1}$?

[wraith dagger]

Ann

[tropic oxide]

that is incorrect.

even if you meant $\frac{1}{2\sqrt{x^2 - 1}}$, that is still incorrect.

[wraith dagger]

Ann

[tropic oxide]

the correct answer is $\frac{x}{\sqrt{x^2 - 1}}$, and in order to get to it you will need to recall the chain rule.